A resistance (R = 12 omega), an inductor (L = 2 H) and a capacitor (C = 5 muF) are connected to an AC generator of frequency 50 Hz. Which of the following statement is correct

Options

A. At resonance, the circuit impedance is zero
B. At resonance, the circuit impedance is 12Omega.
C. At resonance, the frequency of the circuit is 1//2
D. The inductive reactance is less than the capacitive reactance

Answer

To solve the problem, let's analyze the given components in the AC circuit: a resistor (R = 12 Ω), an inductor (L = 2 H), and a capacitor (C = 5 μF) connected to an AC generator with a frequency of 50 Hz. We need to determine which statements about the circuit are correct.

### Step 1: Calculate the Resonant Frequency
The resonant frequency \( f_0 \) for an RLC circuit is given by the formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
Where:
- \( L = 2 \, \text{H} \)
- \( C = 5 \, \mu\text{F} = 5 \times 10^{-6} \, \text{F} \)

Calculating \( LC \):
\[ LC = 2 \times 5 \times 10^{-6} = 10 \times 10^{-6} = 10^{-5} \, \text{H}\cdot\text{F} \]

Now, substituting into the resonant frequency formula:
\[ f_0 = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{1}{2\pi \times 10^{-2.5}} \approx \frac{1}{0.000628} \approx 159.15 \, \text{Hz} \]

### Step 2: Compare Resonant Frequency with Given Frequency
The given frequency \( f = 50 \, \text{Hz} \) is much less than the resonant frequency \( f_0 \approx 159.15 \, \text{Hz} \). Since \( f \neq f_0 \), the circuit is not at resonance.

### Step 3: Calculate Inductive Reactance \( X_L \)
Inductive reactance \( X_L \) is given by:
\[ X_L = 2\pi f L \]
Substituting the values:
\[ X_L = 2\pi \times 50 \times 2 = 200\pi \approx 628.32 \, \Omega \]

### Step 4: Calculate Capacitive Reactance \( X_C \)
Capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{2\pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2\pi \times 50 \times 5 \times 10^{-6}} = \frac{1}{0.000314} \approx 3183.1 \, \Omega \]

### Step 5: Determine the Impedance \( Z \)
The impedance \( Z \) in an RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{12^2 + (628.32 - 3183.1)^2} \]
Calculating \( (X_L - X_C) \):
\[ X_L - X_C = 628.32 - 3183.1 = -2554.78 \]
Now substituting back:
\[ Z = \sqrt{144 + (-2554.78)^2} \approx \sqrt{144 + 6533200.96} \approx \sqrt{6533344.96} \approx 2555.1 \, \Omega \]

### Conclusion
1. The circuit is not at resonance since \( f \neq f_0 \).
2. The calculated impedance \( Z \approx 2555.1 \, \Omega \).